What is the integral of 2x(x^2+6)^9 ?

The integral of 2x(x^2+6)^9 is 1/10 (x^2+6)^{10}+C

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integral of 2x(x^2+6)^9

\int 2 x (x^{2} + 6)^{9} d x

\frac{1}{10} (x^{2} + 6)^{10} + C

Solution steps

\int 2 x (x^{2} + 6)^{9} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int x (x^{2} + 6)^{9} d x

Apply u-substitution

: \int \frac{u ^{9}}{2} d u

= 2 \cdot \int \frac{u ^{9}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int u^{9} d u

Apply the Power Rule

: \frac{u ^{10}}{10}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{10}}{10}

Substitute back

u = x^{2} + 6

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 6 ) ^{10}}{10}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 6 ) ^{10}}{10} : \frac{1}{10} (x^{2} + 6)^{10}

= \frac{1}{10} (x^{2} + 6)^{10}

Add a constant to the solution

= \frac{1}{10} (x^{2} + 6)^{10} + C

\frac{d}{d x} (arcsin (4 - 3 x))

\int_{0}^{\frac{1}{2}} \frac{3}{1 + 4 x ^{2}} d x

\frac{d}{d x} (\frac{( x - 5 x x )}{x})

t^{2} \frac{d y}{d t} + t y = 6

n = 0 \sum \infty \frac{n}{3} (\frac{2}{3})^{n}

What is the integral of 2x(x^2+6)^9 ?
The integral of 2x(x^2+6)^9 is 1/10 (x^2+6)^{10}+C